How to derive the line element from the Fubini-Study metric? On the Wikipedia page on the Fubini-Study metric, the Fubini-Study distance is given (for a Hilbert space) as
$$\gamma(\psi,\phi)=\arccos\sqrt{\frac{\langle \psi|\phi\rangle\langle\phi|\psi\rangle}{\langle\psi|\psi\rangle\langle\phi|\phi\rangle}},$$
and the page says that the infitesimal form can be obtained by setting $\phi=\psi+\delta\psi$, to get $$ds^2=\frac{\langle\delta\psi|\delta\psi\rangle}{\langle\psi|\psi\rangle}-\frac{\langle\delta\psi|\psi\rangle\langle\psi|\delta\psi\rangle}{\langle\psi|\psi\rangle^2}.$$
How does the derivation work? I tried to use a Taylor series approximation of arccos, which cancelled the square root, but I don't know where the rest came from.
 A: Start by considering the following identities, obtained via basic complex algebra and Taylor expansions:
$$|\langle \psi,\psi+\delta\psi\rangle|^2
= \|\psi\|^4\left\lvert 1+ \frac{\langle\psi,\delta\psi\rangle}{\|\psi\|^2}\right\rvert^2 
= \|\psi\|^4 \left(1 +\frac{2\operatorname{Re}\langle\psi,\delta\psi\rangle}{\|\psi\|^2} + \frac{|\langle\psi,\delta\psi\rangle|^2}{\|\psi\|^4}\right)$$
$$
\|\psi+\delta\psi\|^2=\|\psi\|^2 + \|\delta\psi\|^2 + 2\operatorname{Re}\langle\psi,\delta\psi\rangle
$$
$$
\frac{1}{\|\psi+\delta\psi\|^2}\simeq\frac{1}{\|\psi\|^2} \left(
1
- \frac{2\operatorname{Re}\langle\psi,\delta\psi\rangle}{\|\psi\|^2}
- \frac{\|\delta\psi\|^2}{\|\psi\|^2} + \frac{4[\operatorname{Re}\langle\psi,\delta\psi\rangle]^2}{\|\psi\|^4}
\right)
$$
$$
\arccos\sqrt{1-x} \simeq \sqrt x, \quad\text{for } x\to 0^+.
$$

Assuming $\|\psi\|^2=1$ for simplicity (we can keep the $\|\psi\|$ terms around but nothing substantial changes in the expressions) we thus have
$$
\frac{\lvert\langle\psi,\psi+\delta\psi\rangle\rvert^2}{\|\psi+\delta\psi\|^2}
\simeq 1 + \lvert\langle\psi,\delta\psi\rangle\rvert^2 - \|\delta\psi\|^2,
$$
therefore
$$
\gamma(\psi,\psi+\delta\psi)\simeq\arccos\sqrt{1+\lvert\langle\psi,\delta\psi\rangle\rvert^2-\|\delta\psi\|^2}
\simeq \sqrt{\|\delta\psi\|^2-\lvert\langle\psi,\delta\psi\rangle\rvert^2},
$$
and finally
$$
ds^2 \equiv \gamma(\psi,\psi+\delta\psi)^2 \simeq \|\delta\psi\|^2-\lvert\langle\psi,\delta\psi\rangle\rvert^2,
$$
which is the expression we were looking for.
